MGF of a probability function

Question

The MGF of p(Y=y) = Σ(from 0 to y) c(n,x) * p^x * (1-p)^(n-x) + c(n-y,y-x) * p(y-x) * (1-p)^(n-x).

Answer 1

You do not need an MGF. Pull all of the powers to $n$ and as much of those to $y$ out of the sum as possible.

$$\begin{align}P(Y=y) &= \sum_{x= 0}^y \binom n x ~ p^x ~ (1-p)^{n-x} ~ \binom {n-x}{y-x} ~ p^{{y-x}} ~ (1-p)^{n-y}\\[1ex]&=p^y~(1-p)^{2n-y}~\sum_{x=0}^y\frac{n!}{x!~(y-x)!~(n-y)!}~(1-p)^{-x}\\[1ex]&=\binom{n}{y}p^y~(1-p)^{2(n-y)}~\sum_{x=0}^y\binom{y}{x}~(1-p)^{y-x}\end{align}$$

Now recall the binomial expansion: $$\sum_{x=0}^y\binom yx~a^{x}~b^{y-x}=(a+b)^y$$